A pair of equations is shown below:
1 answer:
Remark
The very easiest way to do this is by substitution. The ys are isolated on the left. When you equate what's on the right, you can solve this for 1 variable, the xs.
Substitute and solve
8x - 9 = 4x -1 Subtract 4x from both sides.
8x - 4x - 9 = - 1
4x - 9 = - 1 Add 9 to both sides.
4x = - 1 + 9
4x = 8 Divide by 4
x = 8/4
x = 2
Solve for y
Use either equation to solve for y.
y = 4x - 1
y = 4(2) - 1
y = 8 - 1
y = 7
Answer
The equations intersect at (2,7)
Part B
See the graph below. Notice the intersect point.
The very easiest way to do this is by substitution. The ys are isolated on the left. When you equate what's on the right, you can solve this for 1 variable, the xs.
Substitute and solve
8x - 9 = 4x -1 Subtract 4x from both sides.
8x - 4x - 9 = - 1
4x - 9 = - 1 Add 9 to both sides.
4x = - 1 + 9
4x = 8 Divide by 4
x = 8/4
x = 2
Solve for y
Use either equation to solve for y.
y = 4x - 1
y = 4(2) - 1
y = 8 - 1
y = 7
Answer
The equations intersect at (2,7)
Part B
See the graph below. Notice the intersect point.
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Given:
g(x) = (1/3)x + 2
Part (a)
To find the inverse:
Set y = g(x) = (1/3)x + 2
Swap x and y.
x = (1/3)y + 2.
Solve for y.
(1/3)y = x - 2
y = 3(x - 2).
Set g⁻¹(x) to y.
Answer: g⁻¹(x) = 3(x - 2)
Part (b)
Create the table shown below to graph g(x) and g⁻¹(x).
x g(x) g⁻¹(x)
---- --------- ---------
-8 - 2/3 - 30
-6 0 - 24
-4 2/3 - 18
-2 4/3 - 12
0 2 - 6
2 8/3 0
4 4/3 6
6 4 12
8 14/3 18
Note that when x = -6, g(x) = 0, so that (-6, 0) lies on he black liine.
Therefore the inverse function should yield (0, -6) to be correct. This is so, so g⁻¹ is correct.
Both g(x) and g⁻¹(x)satisfy the vertical line test, so both are functions.
Part (c)
Algebraically, we know that g⁻¹(x) is correct if g(g⁻¹(x)) = x
Use function composition to obtain
g(g⁻¹(x)) = (1/3)*(3x - 6) + 2
= x - 2 + 2
= x
Therefore g⁻¹(x) is correct.
g(x) = (1/3)x + 2
Part (a)
To find the inverse:
Set y = g(x) = (1/3)x + 2
Swap x and y.
x = (1/3)y + 2.
Solve for y.
(1/3)y = x - 2
y = 3(x - 2).
Set g⁻¹(x) to y.
Answer: g⁻¹(x) = 3(x - 2)
Part (b)
Create the table shown below to graph g(x) and g⁻¹(x).
x g(x) g⁻¹(x)
---- --------- ---------
-8 - 2/3 - 30
-6 0 - 24
-4 2/3 - 18
-2 4/3 - 12
0 2 - 6
2 8/3 0
4 4/3 6
6 4 12
8 14/3 18
Note that when x = -6, g(x) = 0, so that (-6, 0) lies on he black liine.
Therefore the inverse function should yield (0, -6) to be correct. This is so, so g⁻¹ is correct.
Both g(x) and g⁻¹(x)satisfy the vertical line test, so both are functions.
Part (c)
Algebraically, we know that g⁻¹(x) is correct if g(g⁻¹(x)) = x
Use function composition to obtain
g(g⁻¹(x)) = (1/3)*(3x - 6) + 2
= x - 2 + 2
= x
Therefore g⁻¹(x) is correct.